Question

Resolva a equaçao matricial.

Answer 1

$\Large\boxed{\begin{array}{l}\begin{bmatrix}\sf1&\sf1&\sf1\\\sf2&\sf1&\sf3\\\sf1&\sf-1&\sf4\end{bmatrix}\cdot\begin{bmatrix}\sf x\\\sf y\\\sf z\end{bmatrix}=\begin{bmatrix}\sf 0\\\sf1\\\sf1\end{bmatrix}\\\begin{bmatrix}\sf 1\cdot x+1\cdot y+1\cdot z\\\sf 2\cdot x+1\cdot y+3\cdot z\\\sf 1\cdot x-1\cdot y+4\cdot z\end{bmatrix}=\begin{bmatrix}\sf0\\\sf1\\\sf1\end{bmatrix}\\\begin{bmatrix}\sf x+y+z\\\sf 2x+y+3z\\\sf x-y+4z \end{bmatrix}=\begin{bmatrix}\sf0\\\sf1\\\sf1\end{bmatrix}\end{array}}$

$\Large\boxed{\begin{array}{l}\sf A=\begin{bmatrix}\sf1&\sf1&\sf1\\\sf2&\sf1&\sf3\\\sf1&\sf-1&\sf4\end{bmatrix}\\\\\sf det~A=1\cdot(4+3)-1\cdot(8-3)+1\cdot(-2-1)\\\sf det~A=7-5-3\\\sf det~A=-1\\\\\sf A_x=\begin{bmatrix}\sf0&\sf1&\sf1\\\sf1&\sf1&\sf3\\\sf1&\sf-1&\sf4\end{bmatrix}\\\sf det~A_x=0\cdot(4+3)-1\cdot(4-3)+1\cdot(-1-2)\\\sf det~A_x=-1-3\\\sf det~A_x=-4\end{array}}$

$\Large\boxed{\begin{array}{l}\sf A_y=\begin{bmatrix}\sf1&\sf0&\sf1\\\sf2&\sf1&\sf3\\\sf1&\sf1&\sf4\end{bmatrix}\\\sf det~A_y=1\cdot(4-3)-0\cdot(8-3)+1\cdot(2-1)\\\sf det~A_y=1+1\\\sf det~A_y=2\\\sf A_z=\begin{bmatrix}\sf1&\sf1&\sf0\\\sf2&\sf1&\sf1\\\sf1&\sf-1&\sf1\end{bmatrix}\\\sf det~A_z=1\cdot(1+1)-1\cdot(2-1)+0\cdot(-2-1)\\\sf det~A_z=2-1\\\sf det~A_z=1\end{array}}$

$\Large\boxed{\begin{array}{l}\sf x=\dfrac{det~A_x}{A}\\\\\sf x=\dfrac{-4}{-1}\\\\\sf x=4\\\sf y=\dfrac{det~A_y}{det~A}\\\\\sf y=\dfrac{2}{-1}\\\\\sf y=-2\\\\\sf z=\dfrac{det~A_z}{det~A}\\\\\sf z=\dfrac{1}{-1}\\\\\sf z=-1\\\sf S=\{4,-2,-1\}\blue{\checkmark}\end{array}}$